Question

A sample of pure alumina hydrate was obtained. A 5.000 g sample of the material was heated carefully in a vacuum oven until no more mass was lost from the sample. After heating, the final weight of the material was 2.6763 g. What was the formula of the hydrated alumina, Al2O3•xH2O? (Enter a whole number for "x") (mol. wt. Al2O3 = 101.96)

Answer 1

Answer: The formula of the hydrated alumina is  $Al_2O_3.5H_2O$

Explanation:

Decomposition of hydrated alumina is given by:

$Al_2O_3.xH_2O\rightarrow Al_2O_3+xH_2O$

Molar mass of $Al_2O_3$ = 101.96 g/mol

According to stoichiometry:

(101.96+18x) g of  $Al_2O_3.xH_2O$ decomposes to give 101.96 g of

$Al_2O_3$

Thus 5.000 g of   $Al_2O_3.xH_2O$ decomposes to give=$\frac {101.96}{(101.96+18x)}\times 5.000$ of H_2O

But it is given 5.000 g of a sample of hydrated salt $Al_2O_3.xH_2O$ was found to contain 2.6763 g of unhydrated salt

Thus we can equate the two equations:

$\frac{101.96}{(101.96+18x)}\times 5.000=2.6763$

$x=5$

Thus the formula of the hydrated alumina is  $Al_2O_3.5H_2O$